\(\int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) (4-2 \log (-\frac {x}{-1+\log (5)}))-4 \log (-\frac {x}{-1+\log (5)})+\log ^2(-\frac {x}{-1+\log (5)})}{x^2} \, dx\) [1253]

Optimal result

Integrand size = 61, antiderivative size = 32 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x \left (x-\frac {\left (2+x+\log (x)-\log \left (\frac {x^2}{x-x \log (5)}\right )\right )^2}{x^2}\right ) \]

[Out]

x*(x-(x-ln(x^2/(-x*ln(5)+x))+2+ln(x))^2/x^2)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(32)=64\).

Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.34, number of steps used = 13, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {14, 2341, 2342, 2413} \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^2-x-\frac {10}{x}-\frac {\log ^2(x)}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x}-\frac {6 \log (x)}{x}+\frac {2 (\log (x)+3)}{x}+\frac {2 (\log (x)+2) \log \left (\frac {x}{1-\log (5)}\right )}{x} \]

[In]

Int[(4 - x^2 + 2*x^3 + Log[x]^2 + Log[x]*(4 - 2*Log[-(x/(-1 + Log[5]))]) - 4*Log[-(x/(-1 + Log[5]))] + Log[-(x
/(-1 + Log[5]))]^2)/x^2,x]

[Out]

-10/x - x + x^2 - (6*Log[x])/x - Log[x]^2/x + (2*(3 + Log[x]))/x + (2*(2 + Log[x])*Log[x/(1 - Log[5])])/x - Lo
g[x/(1 - Log[5])]^2/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2413

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4-x^2+2 x^3+4 \log (x)+\log ^2(x)}{x^2}-\frac {2 (2+\log (x)) \log \left (-\frac {x}{-1+\log (5)}\right )}{x^2}+\frac {\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2}\right ) \, dx \\ & = -\left (2 \int \frac {(2+\log (x)) \log \left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx\right )+\int \frac {4-x^2+2 x^3+4 \log (x)+\log ^2(x)}{x^2} \, dx+\int \frac {\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx \\ & = \frac {2 \log \left (\frac {x}{1-\log (5)}\right )}{x}+\frac {2 (2+\log (x)) \log \left (\frac {x}{1-\log (5)}\right )}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x}+2 \int \frac {-3-\log (x)}{x^2} \, dx+2 \int \frac {\log \left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx+\int \left (\frac {4-x^2+2 x^3}{x^2}+\frac {4 \log (x)}{x^2}+\frac {\log ^2(x)}{x^2}\right ) \, dx \\ & = \frac {2 (3+\log (x))}{x}+\frac {2 (2+\log (x)) \log \left (\frac {x}{1-\log (5)}\right )}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x}+4 \int \frac {\log (x)}{x^2} \, dx+\int \frac {4-x^2+2 x^3}{x^2} \, dx+\int \frac {\log ^2(x)}{x^2} \, dx \\ & = -\frac {4}{x}-\frac {4 \log (x)}{x}-\frac {\log ^2(x)}{x}+\frac {2 (3+\log (x))}{x}+\frac {2 (2+\log (x)) \log \left (\frac {x}{1-\log (5)}\right )}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x}+2 \int \frac {\log (x)}{x^2} \, dx+\int \left (-1+\frac {4}{x^2}+2 x\right ) \, dx \\ & = -\frac {10}{x}-x+x^2-\frac {6 \log (x)}{x}-\frac {\log ^2(x)}{x}+\frac {2 (3+\log (x))}{x}+\frac {2 (2+\log (x)) \log \left (\frac {x}{1-\log (5)}\right )}{x}-\frac {\log ^2\left (\frac {x}{1-\log (5)}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=-\frac {4+x^2-x^3+\log ^2(x)-2 \log (x) \left (-2+\log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x} \]

[In]

Integrate[(4 - x^2 + 2*x^3 + Log[x]^2 + Log[x]*(4 - 2*Log[-(x/(-1 + Log[5]))]) - 4*Log[-(x/(-1 + Log[5]))] + L
og[-(x/(-1 + Log[5]))]^2)/x^2,x]

[Out]

-((4 + x^2 - x^3 + Log[x]^2 - 2*Log[x]*(-2 + Log[-(x/(-1 + Log[5]))]) - 4*Log[-(x/(-1 + Log[5]))] + Log[-(x/(-
1 + Log[5]))]^2)/x)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(67\) vs. \(2(32)=64\).

Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12

method	result	size
parallelrisch	\(-\frac {8-2 x^{3}+2 x^{2}+2 \ln \left (x \right )^{2}-4 \ln \left (x \right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )+2 \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )^{2}+8 \ln \left (x \right )-8 \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )}{2 x}\)	\(68\)
default	\(-\frac {\ln \left (-x \right )^{2}}{x}+\frac {2 \ln \left (-x \right )}{x}-\frac {4}{x}-\frac {\ln \left (\ln \left (5\right )-1\right )^{2}}{x}+2 \ln \left (\ln \left (5\right )-1\right ) \left (\frac {\ln \left (-x \right )}{x}+\frac {1}{x}\right )+\frac {\ln \left (x \right )^{2}}{x}-2 \left (\ln \left (-x \right )-\ln \left (x \right )\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+2 \ln \left (\ln \left (5\right )-1\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x^{2}-x -\frac {2 \ln \left (x \right )}{x}-\frac {4 \ln \left (\ln \left (5\right )-1\right )}{x}\)	\(133\)
risch	\(x^{2}-x +\frac {-4+\pi ^{2}+2 i \ln \left (\ln \left (5\right )-1\right ) \pi \operatorname {csgn}\left (i x \right )^{3}-2 i \ln \left (\ln \left (5\right )-1\right ) \pi \operatorname {csgn}\left (i x \right )^{2}+2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3}+\pi ^{2} \operatorname {csgn}\left (i x \right )^{6}+2 i \pi \ln \left (\ln \left (5\right )-1\right )+4 i \pi -\ln \left (\ln \left (5\right )-1\right )^{2}+\pi ^{2} \operatorname {csgn}\left (i x \right )^{4}-2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{5}-4 \ln \left (\ln \left (5\right )-1\right )+4 i \pi \operatorname {csgn}\left (i x \right )^{3}-4 i \pi \operatorname {csgn}\left (i x \right )^{2}-2 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2}}{x}\)	\(158\)
parts	\(\frac {\left (-\ln \left (5\right )+1\right ) \left (\frac {\left (\ln \left (5\right )-1\right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )^{2}}{x}+\frac {2 \left (\ln \left (5\right )-1\right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )}{x}+\frac {2 \ln \left (5\right )-2}{x}\right )}{\left (\ln \left (5\right )-1\right )^{2}}+\frac {\ln \left (x \right )^{2}}{x}-2 \left (\ln \left (-x \right )-\ln \left (x \right )\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+2 \ln \left (\ln \left (5\right )-1\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+x^{2}-x -\frac {6}{x}-\frac {2 \ln \left (x \right )}{x}-\frac {4 \left (-\ln \left (5\right )+1\right ) \left (\frac {\left (\ln \left (5\right )-1\right ) \ln \left (-\frac {x}{\ln \left (5\right )-1}\right )}{x}+\frac {\ln \left (5\right )-1}{x}\right )}{\left (\ln \left (5\right )-1\right )^{2}}\)	\(175\)

[In]

int((ln(x)^2+(-2*ln(-x/(ln(5)-1))+4)*ln(x)+ln(-x/(ln(5)-1))^2-4*ln(-x/(ln(5)-1))+2*x^3-x^2+4)/x^2,x,method=_RE
TURNVERBOSE)

[Out]

-1/2/x*(8-2*x^3+2*x^2+2*ln(x)^2-4*ln(x)*ln(-x/(ln(5)-1))+2*ln(-x/(ln(5)-1))^2+8*ln(x)-8*ln(-x/(ln(5)-1)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=\frac {x^{3} - x^{2} - \log \left (-\log \left (5\right ) + 1\right )^{2} - 4 \, \log \left (-\log \left (5\right ) + 1\right ) - 4}{x} \]

[In]

integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2-4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^
2,x, algorithm="fricas")

[Out]

(x^3 - x^2 - log(-log(5) + 1)^2 - 4*log(-log(5) + 1) - 4)/x

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} - x + \frac {-4 - \log {\left (-1 + \log {\left (5 \right )} \right )}^{2} - 4 \log {\left (-1 + \log {\left (5 \right )} \right )} + \pi ^{2} + 2 i \pi \log {\left (-1 + \log {\left (5 \right )} \right )} + 4 i \pi }{x} \]

[In]

integrate((ln(x)**2+(-2*ln(-x/(ln(5)-1))+4)*ln(x)+ln(-x/(ln(5)-1))**2-4*ln(-x/(ln(5)-1))+2*x**3-x**2+4)/x**2,x
)

[Out]

x**2 - x + (-4 - log(-1 + log(5))**2 - 4*log(-1 + log(5)) + pi**2 + 2*I*pi*log(-1 + log(5)) + 4*I*pi)/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (34) = 68\).

Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.75 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} + 2 \, {\left (\frac {\log \left (-\frac {x}{\log \left (5\right ) - 1}\right )}{x} + \frac {1}{x}\right )} \log \left (x\right ) - x - \frac {\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 2}{x} - \frac {\log \left (-\frac {x}{\log \left (5\right ) - 1}\right )^{2} + 2 \, \log \left (-\frac {x}{\log \left (5\right ) - 1}\right ) + 2}{x} + \frac {2 \, {\left (\log \left (x\right ) - \log \left (-\log \left (5\right ) + 1\right ) + 2\right )}}{x} - \frac {4 \, \log \left (x\right )}{x} + \frac {4 \, \log \left (-\frac {x}{\log \left (5\right ) - 1}\right )}{x} - \frac {4}{x} \]

[In]

integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2-4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^
2,x, algorithm="maxima")

[Out]

x^2 + 2*(log(-x/(log(5) - 1))/x + 1/x)*log(x) - x - (log(x)^2 + 2*log(x) + 2)/x - (log(-x/(log(5) - 1))^2 + 2*
log(-x/(log(5) - 1)) + 2)/x + 2*(log(x) - log(-log(5) + 1) + 2)/x - 4*log(x)/x + 4*log(-x/(log(5) - 1))/x - 4/
x

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x^{2} - x - \frac {-4 i \, \pi - \pi ^{2} - 2 i \, \pi \log \left (\log \left (5\right ) - 1\right ) + \log \left (\log \left (5\right ) - 1\right )^{2} + 4 \, \log \left (\log \left (5\right ) - 1\right ) + 4}{x} \]

[In]

integrate((log(x)^2+(-2*log(-x/(log(5)-1))+4)*log(x)+log(-x/(log(5)-1))^2-4*log(-x/(log(5)-1))+2*x^3-x^2+4)/x^
2,x, algorithm="giac")

[Out]

x^2 - x - (-4*I*pi - pi^2 - 2*I*pi*log(log(5) - 1) + log(log(5) - 1)^2 + 4*log(log(5) - 1) + 4)/x

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {4-x^2+2 x^3+\log ^2(x)+\log (x) \left (4-2 \log \left (-\frac {x}{-1+\log (5)}\right )\right )-4 \log \left (-\frac {x}{-1+\log (5)}\right )+\log ^2\left (-\frac {x}{-1+\log (5)}\right )}{x^2} \, dx=x\,\left (x-1\right )-\frac {{\left (\ln \left (\ln \left (5\right )-1\right )-\ln \left (-x\right )+\ln \left (x\right )+2\right )}^2}{x} \]

[In]

int((log(x)^2 - 4*log(-x/(log(5) - 1)) - log(x)*(2*log(-x/(log(5) - 1)) - 4) + log(-x/(log(5) - 1))^2 - x^2 +
2*x^3 + 4)/x^2,x)

[Out]

x*(x - 1) - (log(log(5) - 1) - log(-x) + log(x) + 2)^2/x